Question

A rain drop while falling through a cloud made up of stationary particles gains mass at a rate proportional to its instantaneous momentum. If $m$ is its instantaneous mass, $v$ is its instantaneous velocity and $c$ is the constant of proportionality then

• A. $\frac{dv}{dt}=c(\frac{g}{c}-v^2)$
• B. $\frac{dv}{dt}=2c(\frac{g}{c}-2v^2)$
• C. $\frac{dv}{dt}=2c(\frac{g}{c}-v^2)$
• D. $\frac{dv}{dt}=(\frac{g}{c}-2v^{2}c)$

Answer 1

The correct option is A $\frac{dv}{dt}=c(\frac{g}{c}-v^2)$

From the question, $\frac{dm}{dt}\propto mv$
$\frac{dm}{dt}=cmv$

According to Newton's 2nd law,
$m\frac{dv}{dt}+v\frac{dm}{dt}=ma$

Drop is falling , we can say that , $a=g$

$\Rightarrow m\frac{dv}{dt}+v\frac{dm}{dt}=mg$

$\Rightarrow \frac{dv}{dt}=g-\frac{v}{m}\left(\frac{dm}{dt}\right)$

$\Rightarrow \frac{dv}{dt}=g-\frac{v}{m}\left(cmv\right)$

$\Rightarrow \frac{dv}{dt}=g-c{v}^2$

$\Rightarrow \frac{dv}{dt}=c(\frac{g}{c}-v^2)$

Hence, option $\left(a\right)$ is correct.